Ex.7.3 Q2 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

In each of the following find the value of ‘\(k\)’, for which the points are collinear.

(i) \((7, -2), (5, 1), (3, -k)\)

(ii) \((8, 1), (k, -4), (2, -5)\)

 Video Solution
Coordinate Geometry
Ex 7.3 | Question 2

Text Solution

Reasoning:

Three or more points are said to be collinear if they lie on a single straight line .

What is Known:

The \(x\) and \(y\) co-ordinates of the points.

What is  Unknown:

The value of ‘\(k\)’, for which the points are collinear.

Solution:

(i) Given,

  • Let A \((x_1, y_1) = (7, -2)\)
  • Let B \((x_2, y_2) = (5 , 1)\)
  • Let C \((x_3, y_3) = (3, -k)\)

\[\begin{align} & =\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) \\ + x_2 \left( y_3 - y_1 \right) \\ + x_3 \left( y_1 - y_2 \right) \end{bmatrix} \\ &= 0\end{align}\]

(For Collinear Points)

By substituting the values of vertices, \(A, B, C\) in the Equation \((1)\),

\[\begin{align}\frac{1}{2} \begin{bmatrix} 7(1 \! - \! k) \! + \! 5(k \! - \! ( - 2)) \! + \! \\ 3(( - 2) \! - \! 1) \end{bmatrix} \! = \! 0 \\ \begin{bmatrix} 7 \! - \! 7k \! + \! 5k \! + \! 10 \! - \! 9 \end{bmatrix} \! = \! 0 \\ - 2k \! + \! 8 \! = \! 0 \\ k \! = \! 4 \end{align}\]

Hence the given points are collinear for \(\begin{align}k = 4\end{align}\)

(ii) Given,

  • Let A \((x_1, y_1) = (8, 1)\)
  • Let B \((x_2, y_2) = (k , -4)\)
  • Let C \((x_3, y_3) = (2, -5)\)

\[\begin{align} & = \! \frac{1}{2} \! \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) \! + \! \\ x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \\ &= \! 0 \end{align}\]

(For Collinear Points)

By substituting the values of vertices, \(A, B, C\) in the Equation \((1)\),

\[\begin{align} {\frac{1}{2}} \!\begin{bmatrix} 8(- 4 \! - \! ( - 5)) \! + \! k(( - 5) \! - \! (1)) \\ + 2(1 \! - \! ( - 4)) \end{bmatrix} & \! = \! 0\\ 8 \! - \! 6k \! + \! 10 & \! = \! 0 \\ \left( {\text{By Transposing}} \right)\\ 6k & \! = \!  18\\ k & \! = \!  3\end{align}\]

Hence, the given points are collinear for \(\begin{align}k = 3 \end{align}\)

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